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Hard problem with right triangles  TOPIC_SOLVED

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
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Postby stapel_eliz on Mon Nov 29, 2010 7:54 pm

dade488 wrote:So I get :
tanbeta=(m-s)/((s+m)/tanalpha)

...which simplifies as:

. . . . .

Now cross-multiply, expand, and solve for "m=". :wink:

dade488 wrote:If I keep going I get this:

tantheta= 5/(x+(36/x))...

Please show your work!
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